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Math Help - Sine/ Cosine Rule Problems

  1. #1
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    Sine/ Cosine Rule Problems



    Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was x^2 -(14)(cos50)x -51=0 and i'm not even sure that's right. Some help would be appreciated.

    Thanks!
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  2. #2
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    Quote Originally Posted by ns2583 View Post


    Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was x^2 -(14)(cos50)x -51=0 and i'm not even sure that's right. Some help would be appreciated.

    Thanks!
    Your equation is correct.

    You can solve it for x using the Quadratic formula.
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  3. #3
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    Quote Originally Posted by ns2583 View Post


    Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was x^2 -(14)(cos50)x -51=0 and i'm not even sure that's right. Some help would be appreciated.

    Thanks!
    (a) is a direct application of the cosine rule: 10^2 = 7^2 + x^2 - 2 (x)(7) \cos 50^0.

    (b) \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0} .... (1)

    \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0} .... (1)

    \frac{7}{\sin \beta} = \frac{10}{\sin 50^0} .... (2)

    From (2) get \beta. From \alpha + \beta + 50^0 = 180^0 get \alpha. Now use (1) to solve for x.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Your equation is correct.

    You can solve it for x using the Quadratic formula.
    When I put it in my calc, it said there was a nonreal answer. :S

    Quote Originally Posted by mr fantastic View Post
    (a) is a direct application of the cosine rule:

    (b) \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0} .... (1)

    \frac{x}{\sin \alpha} = \frac{10}{\sin 50^0} .... (1)

    \frac{7}{\sin \beta} = \frac{10}{\sin 50^0} .... (2)

    From (2) get \beta. From \alpha + \beta + 50^0 = 180^0 get \alpha. Now use (1) to solve for x.
    Got it, thanks!
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  5. #5
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    Quote Originally Posted by ns2583 View Post
    Quote Originally Posted by Prove It View Post
    Your equation is correct.

    You can solve it for x using the Quadratic formula.
    When I put it in my calc, it said there was a nonreal answer. :S
    Huh? Just use the quadratic formula:

    \begin{aligned}<br />
x^2 - 14(\cos 50^{\circ}) x - 51 &= 0 \\<br />
x^2 - 9x - 51 &\approx 0 \\<br />
x &\approx \frac{9 \pm \sqrt{(-9)^2 - 4(1)(-51)}}{2(1)} \\<br />
&\approx \frac{9 \pm \sqrt{284.98}}{2} \\<br />
\end{aligned}

    x \approx 12.94
    x \approx -3.94
    Reject the 2nd solution, so x is about 13 cm.


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  6. #6
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    OH, i see. Thanks.
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